Prove $\int_{-\pi}^{\pi}\sin \sin x \,dx=0$ without using the fact that $\sin(x)$ is odd. Prove $$\large\int_{-\pi}^{\pi}\sin (\sin x) \,dx =0$$ without using the fact that $\sin(x)$ is odd.
Computing this in wolfram says that it is uncomputable, which leads me to believe that the only way to find this would be methods for solving definite integrals. I am wondering if it is possible with any other techniques such as DUIS or residues?
 A: $$\int_{-\pi}^\pi \sin(\sin(x)) \; dx = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx$$ (the interchange of sum and integral justified by uniform absolute convergence)
so it suffices to show that $\int_{-\pi}^\pi \sin^{2n+1}\; dx = 0$ for nonnegative integers $n$.
Now using the substitution $u = \cos(x)$, 
$$\int_{-\pi}^\pi \sin^{2n+1}(x)\; dx = \int_{-\pi}^\pi (1 - \cos^2(x))^n \sin(x)\; dx = -\int_{-1}^{-1} (1-u^2)^n \; du = 0 $$
A: Here is a very weird solution. Notice that for $s \in \Bbb{R}$,
$$ \int_{-\pi}^{\pi} \sin(s\sin x) \, dx = \Im \int_{-\pi}^{\pi} \exp(is\sin x) \, dx. $$
Now it follows that
\begin{align*}
\int_{-\pi}^{\pi} \exp(is\sin x) \, dx
&= \int_{-\pi}^{\pi} \exp(se^{ix}/2) \exp(-se^{-ix}/2) \, dx \\
&= \int_{-\pi}^{\pi} \exp(se^{ix}/2) \cdot \overline{\exp(-se^{ix}/2)} \, dx \\
&= 2\pi \sum_{n=0}^{\infty} \left\{ \frac{1}{n!}\left(\frac{s}{2}\right)^{n} \right\}\left\{ \frac{(-1)^{n}}{n!}\left(\frac{s}{2}\right)^{n} \right\} \\
&= 2\pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n!)^{2}}\left(\frac{s}{2}\right)^{2n} \\
&= 2\pi J_{0}(s),
\end{align*}
where $J_{0}$ is the Bessel function of the first kind. Here we only need the fact that $J_{0}(s) \in \Bbb{R}$ if $s \in \Bbb{R}$. Consequently we get
$$ \int_{-\pi}^{\pi} \sin(s\sin x) \, dx = 0 $$
and
$$ \int_{-\pi}^{\pi} \cos(s\sin x) \, dx = 2\pi J_{0}(s). $$
(Of course, now both two identities extend to all of $s \in \Bbb{C}$ by the principle of analytic continuation.)
A: I think this question is a bit funny (and in principal, I think that any calculation leading to the desired conclusion uses that sine is odd in one or another way). 
Are we allowed to use $\sin(t\pm\pi)=-\sin(t)$ and that $\sin t=\frac{\exp(it)-\exp(-it)}{2i}$ (I agree that it follows from the second identity directly that sine is odd. But in some sense it is not worse than saying that $\sin t=\text{Im}\,\exp(it)$ and then use that $\overline{\exp(it)}=\exp(-it)$)? 
Nevertheless, here are some calculations using these formulas, leading to the desired result:
We divide the integral in two pieces
$$
I=\int_{-\pi}^\pi \sin(\sin x)\,dx = \int_{-\pi}^0\sin(\sin x)\,dx+\int_0^{\pi}\sin(\sin x)\,dx.
$$
Performing the change of variables $t=x+\pi$ and $t=x-\pi$ respectively in these integrals give
$$
\begin{align}
I&=\int_0^{\pi}\sin(\sin(t-\pi))\,dt+\int_{-\pi}^0 \sin(\sin(t+\pi))\,dt\\
&=\int_{-\pi}^{\pi} \sin(-\sin t)\,dt.
\end{align}
$$
Next, we use that $\sin z=\frac{\exp(iz)-\exp(-iz)}{2i}$,
$$
\begin{align}
I & = \int_{-\pi}^\pi \sin(\sin x)\,dx\\
 & = \int_{-\pi}^{\pi}\frac{\exp(i\sin x)-\exp(-i\sin x)}{2i}\,dx\\
 & = -\int_{-\pi}^{\pi} \frac{\exp(-i\sin x)-\exp(i\sin x)}{2i}\,dx\\
 & = -\int_{-\pi}^{\pi} \sin(-\sin x)\,dx\\
 & = -I.
\end{align}
$$
Thus $I=0$.
A: Another silly answer, using complex analytic methods (similar to sos440's answer, but avoiding use of Bessel functions):
Rewrite the integrand using Euler's formulas and put $z = e^{ix}$, thus mapping $[-\pi,\pi]$ to the unit circle (some algebra omitted):
$$
\int_{-\pi}^\pi \sin \sin x \, dx =
-\frac1{2} \int_{|z|=1}  \frac{\exp\Big( \frac12 ( z - \frac1z ) \Big) - \exp\Big( \frac12 ( -z + \frac1z ) \Big)}{z}\,dz.
$$
The integrand has an essential singularity at $z=0$, but we can still compute the relevant residue. Thanks to the $z$ in the denominator, we only have to compute the $0$:th terms of the Laurent series for the numerator.
We have
\begin{align}
\exp\Big( \frac12 ( z - \frac1z ) \Big) &=
e^{z/2} \cdot e^{-1/(2z)} \\
&= \Big( 1 + \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big)
\Big( 1 - \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big)
\end{align}
Hence, the $0$:th term will be
$$
1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots
$$
Similarly
\begin{align}
\exp\Big( \frac12 (-z + \frac1z ) \Big) &=
e^{-z/2} \cdot e^{1/(2z)} \\
&= \Big( 1 - \frac{1}{1!} \big( \frac{z}{2} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 - \cdots \Big)
\Big( 1 + \frac{1}{1!} \big( \frac{1}{2z} \big) + \frac{1}{2!} \big( \frac{z}{2} \big)^2 + \cdots \Big)
\end{align}
And again, the $0$:th term will be
$$
1 - \frac{1}{1!} \frac{1}{2^2} + \frac{1}{2!} \frac{1}{2^4} - \frac{1}{3!} \frac{1}{2^6} + \cdots
$$
Summing up, the $0$:th term in the Laurent series for the numerator vanishes, and by the residue theorem, the integral, unsurprisingly is $0$.
Of course, this approach also uses, albeit implicitly, since I bothered to write out a number of unnecessary calculations, that the integrand is odd.
